Question: The polynomial $p(x)=x^3-7x-6$ has a known factor of $(x+1)$. Rewrite $p(x)$ as a product of linear factors. $p(x)=$
Answer: We know $(x+1)$ is a factor of $p(x)$. This means that $p(x)=(x+1)\cdot q(x)$ for some polynomial $q(x)$. We can find $q(x)$ using polynomial division, and then we can factor $q(x)$. This way, we will be able to rewrite $p(x)$ as a product of linear factors. Dividing $p(x)$ by $(x+1)$ Notice that $p(x)$ is missing a $2^{\text{nd}}$ degree term. Let's add it as $0x^2$. $\begin{array}{r} x^2-\phantom{1}x-6 \\ x+1|\overline{x^3+0x^2-7x-6} \\ \mathllap{-(}\underline{x^3+\phantom{0}x^2\phantom{-7x-6}\rlap)} \\ -x^2-7x-6 \\ \mathllap{-(}\underline{-x^2-\phantom{7}x\phantom{-6}\rlap)} \\ -6x-6 \\ \mathllap{-(}\underline{-6x-6\rlap)} \\ 0 \end{array}$ We find that $q(x)=x^2-x-6$. Factoring $q(x)$ We can factor $x^2{-1}x{-6}$ as $(x+m)(x+n)$ where $m+n={-1}$ and $m\cdot n={-6}$. Such numbers are $2$ and $-3$, so the factored expression is $(x+2)(x-3)$. Putting it all together $\begin{aligned} p(x)&=x^3-7x-6 \\\\ &=(x+1)(x^2-x-6) \\\\ &=(x+1)(x+2)(x-3) \end{aligned}$